Official course description: “Introduction to the concepts used in the modern treatment of solids. The student is assumed to be familiar with elementary quantum mechanics. Topics include: bonding in solids, crystal structures, lattice vibrations, free electron model of metals, band structure, thermal properties, magnetism and superconductivity (time permitting)”

This document contains:

• Plain old lecture notes. These mirror what was covered in class, possibly augmented with additional details.
• Personal notes exploring details that were not clear to me from the lectures, or from the texts associated with the lecture material.
• Assigned problems. Like anything else take these as is.
• Some worked problems attempted as course prep, for fun, or for test preparation, or post test reflection.
• Links to Mathematica workbooks associated with this course.
My thanks go to Professor Julian for teaching this course.

NOTE: This v.5 update of these notes is still really big (~18M). Some of my mathematica generated 3D images result in very large pdfs.

January 07, 2014 Two body harmonic oscillator in 3D
Figure out a general solution for two interacting harmonic oscillators, then use the result to calculate the matrix required for a 2D two atom diamond lattice with horizontal, vertical and diagonal nearest neighbour coupling.

NOTE: This v.2 update of these notes is really big (~18M), despite being only half way into the course. My mathematica generated images appear to result in very large pdfs, and I’m looking at trying pdfsizeopt to reduce the size before posting the next update (or take out the density plots from my problem set 1 solutions).

This set of notes includes the following these additions (not many of which were posted separately for this course)

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

Disclaimer

Question: Bose-Einstein condensation (BEC) in one and two dimensions

Obtain the density of states in one and two dimensions for a particle with an energy-momentum relation

Using this, show that for particles whose number is conserved the BEC transition temperature vanishes in these cases – so we can always pick a chemical potential which preserves a constant density at any temperature.

Answer

We’d like to evaluate

We’ll use

where the roots of are . With

the roots of are

The derivative of evaluated at these roots are

In 2D, we can evaluate over a shell in space

or

In 1D we have

Observe that this time for 1D, unlike in 2D when we used a radial shell in space, we have contributions from both the delta function roots. Our end result is

To consider the question of the BEC temperature, we’ll need to calculate the density. For the 2D case we have

Recall for the 3D case that we had an upper bound as . We don’t have that for this 2D density, so for any value of , a corresponding value of can be found. That is

For the 1D case we have

or

See fig. 1.1 for plots of for , the respective results for the 1D, 2D and 3D densities respectively.

Fig 1.1: Density integrals for 1D, 2D and 3D cases

We’ve found that is also unbounded as , so while we cannot invert this easily as in the 2D case, we can at least say that there will be some for any value of that allows the density (and thus the number of particles) to remain fixed.

Question: Estimating the BEC transition temperature

Find data for the atomic mass of liquid He and its density at ambient atmospheric pressure and hence estimate its BEC temperature assuming interactions are unimportant (even though this assumption is a very bad one!).

For dilute atomic gases of the sort used in Professor
Thywissen’s lab
, one typically has a cloud of atoms confined to an approximate cubic region with linear dimension 1 . Find the density – it is pretty low, so interactions can be assumed to be extremely weak. Assuming these are Rb atoms, estimate the BEC transition temperature.

Answer

With an atomic weight of 4.0026, the mass in grams for one atom of Helium is

With the density of liquid He-4, at 5.2K (boiling point): 125 grams per liter, the number density is

In class the was found to be

So for liquid helium we have

The number density for the gas in Thywissen’s lab is

The mass of an atom of Rb is

which gives us

Question: Phonons in two dimensions

for small momenta , where is the speed of sound. Assuming a two-dimensional crystal, phonons only propagate along the plane containing the atoms. Find the specific heat of this crystal due to phonons at low temperature. Recall that phonons are not conserved, so there is no chemical potential associated with maintaining a fixed phonon density.

around the mean energy . The first derivative part of the expansion is simple enough

The peak energy will be where this derivative equals zero. That is

or

With

We have

so that

So far so good. Reading the text, the expansion of the logarithm of around wasn’t clear. Let’s write that out in full. To two terms that is

The first order term has the derivative of the logarithm of . Since the logarithm is monotonic and the derivative of has been shown to be zero at , this must be zero. We can also see this explicitly by computation

For the second derivative we have

Somehow this is supposed to come out to ? Backing up, we have

I still don’t see how to get out of this? is a derivative with respect to temperature, but here we have derivatives with respect to energy (keeping fixed)?

One can measure the specific heat in this Bose condensation phenomina for materials such as Helium-4 (spin 0). However, it turns out that Helium-4 is actually quite far from an ideal Bose gas.

Photon gas

A system that is much closer to an ideal Bose gas is that of a gas of photons. To a large extent, photons do not interact with each other. This allows us to calculate black body phenomina and the low temperature (cosmic) background radiation in the universe.

An important distinction between a photon sea and some of these other systems is that the photon number is actually not fixed.

Photon numbers are not “conserved”.

If a photon interacts with an atom, it can impart energy and disappear. An excited atom can emit a photon and change its energy level. In a thermodynamic system we can generally expect that introducing heat will generate more photons, whereas a cold sink will tend to generate fewer photons.

We have a few special details that distinguish photons that we’ll have to consider.

spin 1.

massless, moving at the speed of light.

have two polarization states.

Because we do not have a constraint on the number of particles, we essentially have no chemical potential, even in the grand canonical scheme.

Writing

Our number density, since we have no chemical potential, is of the form

Observe that the average number of photons in this system is temperature dependent. Because this chemical potential is not there, it can be quite easy to work out a number of the thermodynamic results.

Photon average energy density

We’ll now calculate the average energy density of the photons. The energy of a single photon is

so that the average energy density is

Mathematica tells us that this integral is

for an end result of

Phonons and other systems

There is a very similar phenomina in matter. We can discuss lattice vibrations in a solid. These are called phonon modes, and will have the same distribution function where the only difference is that the speed of light is replaced by the speed of the sound wave in the solid. Once we understand the photon system, we are able to look at other Bose distributions such as these phonon systems. We’ll touch on this very briefly next time.

Disclaimer

Last time we found that the low temperature behaviour or the chemical potential was quadratic as in fig. 1.1.

Fig 1.1: Fermi gas chemical potential

Specific heat

where

Low temperature

The only change in the distribution fig. 1.2, that is of interest is over the step portion of the distribution, and over this range of interest is approximately constant as in fig. 1.3.

Fig 1.2: Fermi distribution

Fig 1.3: Fermi gas density of states

so that

Here we’ve made a change of variables , so that we have near cancelation of the factor

Here we’ve extended the integration range to since this doesn’t change much. FIXME: justify this to myself? Taking derivatives with respect to temperature we have

With , we have for

Using eq. 1.1.4 at the Fermi energy and

we have

Giving

or

This is illustrated in fig. 1.4.

Fig 1.4: Specific heat per Fermion

Relativisitic gas

Relativisitic gas

graphene

massless Dirac Fermion

Fig 1.5: Relativisitic gas energy distribution

We can think of this state distribution in a condensed matter view, where we can have a hole to electron state transition by supplying energy to the system (i.e. shining light on the substrate). This can also be thought of in a relativisitic particle view where the same state transition can be thought of as a positron electron pair transition. A round trip transition will have to supply energy like as illustrated in fig. 1.6.

Fig 1.6: Hole to electron round trip transition energy requirement

Graphene

Consider graphene, a 2D system. We want to determine the density of states ,